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B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)
B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)
B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x 0
B = 0
B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)
B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)
B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x 0
B = 0
ta thấy vế thứ hai có kết quả bằng 0
=>(1999x1998+1998x1997)x0
chằng cần tìm kết quả mà =>B=0
\(M=1+\frac{1}{199}+1+\frac{2}{198}+1+....+\frac{198}{2}+1=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}\)
\(=200.\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)=200 T
\(S=\frac{T}{200T}=\frac{1}{200}\)
\(\frac{3}{4}+\frac{2}{3}-\frac{1}{6}\)
\(=\frac{3}{4}+\frac{2}{3}\)
\(=\frac{17}{12}\)
\(=\frac{17}{12}-\frac{1}{6}\)
\(=\frac{90}{72}\)
B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)
B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)
B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x 0
B = 0
b) \(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+...+\frac{87}{1000}+\frac{99}{1000}\)
\(=\frac{1+13+25+...+85+97}{1000}=\frac{\left(97+1\right).\left[\left(97-1\right):12+1\right]:2}{1000}\)
\(=\frac{49.9}{1000}=\frac{441}{1000}.\) ( Đề bài sai nhé bạn tử số : 1; 13; 25; 37; 49 ; 61; 73; 85 ; 97. )
Câu c:B = (1999x1998 + 1998x1997)x(1+1/2: 1 1/2 - 11/3)
B = (1999x1998 + 1998x1997)x(1+1/2:3/2-4/3)
B = (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)
B = (1999 x 1998 + 1998 x 1997) x 0
B = 0
\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(A>\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2015-2014}{2014.2015}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(A>1-\frac{1}{2015}\)
Mà \(\frac{1}{2015}< \frac{1}{4}\Rightarrow1-\frac{1}{2015}>1-\frac{1}{4}=\frac{3}{4}\Rightarrow A>\frac{3}{4}\)
(1999 x 1998 + 1998 x 1997) x (1 + 1/2 : 3/2 - 4/3)
= (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x 0
= 0
(1999 x 1998 + 1998 x 1997) x (1 + 1/2 : 3/2 - 4/3)
= (1999x1998 +1998 x1997) x (1+1/2x2/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x(1+1/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x (4/3 - 4/3)
= (1999 x 1998 + 1998 x 1997) x 0
= 0